Question Number 208359 by alcohol last updated on 13/Jun/24
Commented by alcohol last updated on 13/Jun/24
$${please}\:{help} \\ $$
Answered by Berbere last updated on 17/Jun/24
![f(x)=x^n +x−1⇒f′(x)=nx^(n−1) +1≥1;f increase f(0)=−1,f(1)=1 ∃c_n ∣f(x_n )=0 x_n ^n +x−1=0 f_(n+1) (x_n )=x_n ^(n+1) +x_n −1≤x_n ^n +x_n −1=0 x_(n+1) >x_n ⇒x_n cv x_n =l if l<1⇒lim_(n→∞) l^n =0⇒l−1=0⇒l=1 absurd ⇒l=1 since 0≤x_n ≤1 1−x_n =x_n ^n =u_n ⇒u_n →0⇒ 1−x_n =x_n ^n ⇒nln(x_n )−ln(1−x_n )=0 x_n =x_n =1−u_n nln(1−u_n )−ln(u_n )=0⇒u_n solution of f(x)=nln(1−x)−ln(x) u_n ∈]0,1[ f′(x)=((−n)/(1−x))−(1/x)<0 f(((ln(n))/(2n)))=nln(1−((ln(n))/(2n)))−ln(((ln(n))/(2n))) ∼−((ln(n))/2)−ln(ln(n))+ln(2n) =((ln(n))/2)−ln(ln(n))>0 for n enough big f(2((ln(n))/n))=nln(1−((2ln(n))/n))−ln(((2ln(n))/n)) ∼−2ln(n)−ln(2ln(n)+ln(n)∼−ln(n)<0 for n enough big f(((ln(n))/(2n)))>;f(((2ln(n))/n))>0 f(u_n )=0 f decrease ⇒ ((ln(n))/(2n))<u_n <((2ln(n))/n)](https://www.tinkutara.com/question/Q208368.png)
$${f}\left({x}\right)={x}^{{n}} +{x}−\mathrm{1}\Rightarrow{f}'\left({x}\right)={nx}^{{n}−\mathrm{1}} +\mathrm{1}\geqslant\mathrm{1};{f}\:{increase} \\ $$$${f}\left(\mathrm{0}\right)=−\mathrm{1},{f}\left(\mathrm{1}\right)=\mathrm{1}\:\exists{c}_{{n}} \:\mid{f}\left({x}_{{n}} \right)=\mathrm{0} \\ $$$${x}_{{n}} ^{{n}} +{x}−\mathrm{1}=\mathrm{0} \\ $$$${f}_{{n}+\mathrm{1}} \left({x}_{{n}} \right)={x}_{{n}} ^{{n}+\mathrm{1}} +{x}_{{n}} −\mathrm{1}\leqslant{x}_{{n}} ^{{n}} +{x}_{{n}} −\mathrm{1}=\mathrm{0} \\ $$$${x}_{{n}+\mathrm{1}} >{x}_{{n}} \Rightarrow{x}_{{n}} \:{cv}\:{x}_{{n}} ={l} \\ $$$${if}\:{l}<\mathrm{1}\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{l}^{{n}} =\mathrm{0}\Rightarrow{l}−\mathrm{1}=\mathrm{0}\Rightarrow{l}=\mathrm{1}\:{absurd} \\ $$$$\Rightarrow{l}=\mathrm{1}\:{since}\:\mathrm{0}\leqslant{x}_{{n}} \leqslant\mathrm{1} \\ $$$$\mathrm{1}−{x}_{{n}} ={x}_{{n}} ^{{n}} ={u}_{{n}} \Rightarrow{u}_{{n}} \rightarrow\mathrm{0}\Rightarrow \\ $$$$\mathrm{1}−{x}_{{n}} ={x}_{{n}} ^{{n}} \Rightarrow{nln}\left({x}_{{n}} \right)−{ln}\left(\mathrm{1}−{x}_{{n}} \right)=\mathrm{0} \\ $$$${x}_{{n}} ={x}_{{n}} =\mathrm{1}−{u}_{{n}} \\ $$$${nln}\left(\mathrm{1}−{u}_{{n}} \right)−{ln}\left({u}_{{n}} \right)=\mathrm{0}\Rightarrow{u}_{{n}} \:{solution}\:{of}\:{f}\left({x}\right)={nln}\left(\mathrm{1}−{x}\right)−{ln}\left({x}\right) \\ $$$$\left.{u}_{{n}} \in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$${f}'\left({x}\right)=\frac{−{n}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{{x}}<\mathrm{0} \\ $$$${f}\left(\frac{{ln}\left({n}\right)}{\mathrm{2}{n}}\right)={nln}\left(\mathrm{1}−\frac{{ln}\left({n}\right)}{\mathrm{2}{n}}\right)−{ln}\left(\frac{{ln}\left({n}\right)}{\mathrm{2}{n}}\right) \\ $$$$\sim−\frac{{ln}\left({n}\right)}{\mathrm{2}}−{ln}\left({ln}\left({n}\right)\right)+{ln}\left(\mathrm{2}{n}\right) \\ $$$$=\frac{{ln}\left({n}\right)}{\mathrm{2}}−{ln}\left({ln}\left({n}\right)\right)>\mathrm{0}\:{for}\:{n}\:{enough}\:{big} \\ $$$${f}\left(\mathrm{2}\frac{{ln}\left({n}\right)}{{n}}\right)={nln}\left(\mathrm{1}−\frac{\mathrm{2}{ln}\left({n}\right)}{{n}}\right)−{ln}\left(\frac{\mathrm{2}{ln}\left({n}\right)}{{n}}\right) \\ $$$$\sim−\mathrm{2}{ln}\left({n}\right)−{ln}\left(\mathrm{2}{ln}\left({n}\right)+{ln}\left({n}\right)\sim−{ln}\left({n}\right)<\mathrm{0}\right. \\ $$$${for}\:{n}\:{enough}\:{big}\: \\ $$$${f}\left(\frac{{ln}\left({n}\right)}{\mathrm{2}{n}}\right)>;{f}\left(\frac{\mathrm{2}{ln}\left({n}\right)}{{n}}\right)>\mathrm{0}\:{f}\left({u}_{{n}} \right)=\mathrm{0}\:{f}\:{decrease} \\ $$$$\Rightarrow\:\:\frac{{ln}\left({n}\right)}{\mathrm{2}{n}}<{u}_{{n}} <\frac{\mathrm{2}{ln}\left({n}\right)}{{n}} \\ $$